The matrix corresponds to a homomorphism $mathbb{Z^2} → mathbb{Z^3}$
In Aluffi's Algebra in 4.3. Reading a presentation it says:
For example, take $M=begin{pmatrix}1&3\2&3\5&9end{pmatrix}$; this matrix corresponds to a homomorphism $mathbb{Z^2} → mathbb{Z^3}$, hence to a Z-module.
1- What does it mean it corresponds to a homomorphism $mathbb{Z^2} → mathbb{Z^3}$? For example, does it mean $begin{pmatrix}1&3\2&3\5&9end{pmatrix} begin{pmatrix}a\bend{pmatrix} = begin{pmatrix}a'\b'\c'end{pmatrix}$ holds for any $a,b,a',b',c'$?
2- What does it mean it corresponds to a a Z-module? For example, in an R-module M we have a map $R times M to M$ so how $mathbb{Z^2} → mathbb{Z^3}$ instead of $mathbb{Z^2} → mathbb{Z^2}$ if it is a Z-module?
abstract-algebra matrices group-homomorphism
asked Dec 10 '18 at 3:11
72D
562116
add a comment |
In Aluffi's Algebra in 4.3. Reading a presentation it says:
For example, take $M=begin{pmatrix}1&3\2&3\5&9end{pmatrix}$; this matrix corresponds to a homomorphism $mathbb{Z^2} → mathbb{Z^3}$, hence to a Z-module.
1- What does it mean it corresponds to a homomorphism $mathbb{Z^2} → mathbb{Z^3}$? For example, does it mean $begin{pmatrix}1&3\2&3\5&9end{pmatrix} begin{pmatrix}a\bend{pmatrix} = begin{pmatrix}a'\b'\c'end{pmatrix}$ holds for any $a,b,a',b',c'$?
2- What does it mean it corresponds to a a Z-module? For example, in an R-module M we have a map $R times M to M$ so how $mathbb{Z^2} → mathbb{Z^3}$ instead of $mathbb{Z^2} → mathbb{Z^2}$ if it is a Z-module?
abstract-algebra matrices group-homomorphism
asked Dec 10 '18 at 3:11
72D
562116
You seem to have misread the excerpt; the matrix corresponds to a homomorphism $Bbb Z^2 to Bbb Z^3$. Do you understand how this is different from what you typed? Does this clear anything up?
– Omnomnomnom
Dec 10 '18 at 3:30
I'm confused with your use of subscripts vs what I think ought to be superscripts. Usually, $mathbb{Z_2}$ would be the integers mod $2$, where as $mathbb{Z^2}$ would be the direct product of the integers with itself, ie, pairs of integers where operations are done by component. Where you are using matrices, I'm assuming you actually want the latter.
– Josh B.
Dec 10 '18 at 3:31
Also, it is not at all clear what you mean by $$ M=begin{pmatrix}1&3\2&3\5&9end{pmatrix} begin{pmatrix}1\0end{pmatrix} ne begin{pmatrix}1\2\0end{pmatrix} $$
– Omnomnomnom
Dec 10 '18 at 3:34
@72D do you understand the difference between $Bbb Z_2$ and $Bbb Z^2$?
– Omnomnomnom
Dec 10 '18 at 3:38
@Omnomnomnom, I edited but still I have same questions.
– 72D
Dec 10 '18 at 4:12
add a comment |
In Aluffi's Algebra in 4.3. Reading a presentation it says:
For example, take $M=begin{pmatrix}1&3\2&3\5&9end{pmatrix}$; this matrix corresponds to a homomorphism $mathbb{Z^2} → mathbb{Z^3}$, hence to a Z-module.
1- What does it mean it corresponds to a homomorphism $mathbb{Z^2} → mathbb{Z^3}$? For example, does it mean $begin{pmatrix}1&3\2&3\5&9end{pmatrix} begin{pmatrix}a\bend{pmatrix} = begin{pmatrix}a'\b'\c'end{pmatrix}$ holds for any $a,b,a',b',c'$?
2- What does it mean it corresponds to a a Z-module? For example, in an R-module M we have a map $R times M to M$ so how $mathbb{Z^2} → mathbb{Z^3}$ instead of $mathbb{Z^2} → mathbb{Z^2}$ if it is a Z-module?
abstract-algebra matrices group-homomorphism
asked Dec 10 '18 at 3:11
72D
562116
In Aluffi's Algebra in 4.3. Reading a presentation it says:
For example, take $M=begin{pmatrix}1&3\2&3\5&9end{pmatrix}$; this matrix corresponds to a homomorphism $mathbb{Z^2} → mathbb{Z^3}$, hence to a Z-module.
1- What does it mean it corresponds to a homomorphism $mathbb{Z^2} → mathbb{Z^3}$? For example, does it mean $begin{pmatrix}1&3\2&3\5&9end{pmatrix} begin{pmatrix}a\bend{pmatrix} = begin{pmatrix}a'\b'\c'end{pmatrix}$ holds for any $a,b,a',b',c'$?
2- What does it mean it corresponds to a a Z-module? For example, in an R-module M we have a map $R times M to M$ so how $mathbb{Z^2} → mathbb{Z^3}$ instead of $mathbb{Z^2} → mathbb{Z^2}$ if it is a Z-module?
abstract-algebra matrices group-homomorphism
abstract-algebra matrices group-homomorphism
asked Dec 10 '18 at 3:11
72D
562116
asked Dec 10 '18 at 3:11
72D
562116
edited Dec 10 '18 at 3:50
asked Dec 10 '18 at 3:11
72D
562116
asked Dec 10 '18 at 3:11
72D
562116
asked Dec 10 '18 at 3:11
72D
562116
562116
You seem to have misread the excerpt; the matrix corresponds to a homomorphism $Bbb Z^2 to Bbb Z^3$. Do you understand how this is different from what you typed? Does this clear anything up?
– Omnomnomnom
Dec 10 '18 at 3:30
I'm confused with your use of subscripts vs what I think ought to be superscripts. Usually, $mathbb{Z_2}$ would be the integers mod $2$, where as $mathbb{Z^2}$ would be the direct product of the integers with itself, ie, pairs of integers where operations are done by component. Where you are using matrices, I'm assuming you actually want the latter.
– Josh B.
Dec 10 '18 at 3:31
Also, it is not at all clear what you mean by $$ M=begin{pmatrix}1&3\2&3\5&9end{pmatrix} begin{pmatrix}1\0end{pmatrix} ne begin{pmatrix}1\2\0end{pmatrix} $$
– Omnomnomnom
Dec 10 '18 at 3:34
@72D do you understand the difference between $Bbb Z_2$ and $Bbb Z^2$?
– Omnomnomnom
Dec 10 '18 at 3:38
@Omnomnomnom, I edited but still I have same questions.
– 72D
Dec 10 '18 at 4:12
add a comment |
You seem to have misread the excerpt; the matrix corresponds to a homomorphism $Bbb Z^2 to Bbb Z^3$. Do you understand how this is different from what you typed? Does this clear anything up?
– Omnomnomnom
Dec 10 '18 at 3:30
I'm confused with your use of subscripts vs what I think ought to be superscripts. Usually, $mathbb{Z_2}$ would be the integers mod $2$, where as $mathbb{Z^2}$ would be the direct product of the integers with itself, ie, pairs of integers where operations are done by component. Where you are using matrices, I'm assuming you actually want the latter.
– Josh B.
Dec 10 '18 at 3:31
Also, it is not at all clear what you mean by $$ M=begin{pmatrix}1&3\2&3\5&9end{pmatrix} begin{pmatrix}1\0end{pmatrix} ne begin{pmatrix}1\2\0end{pmatrix} $$
– Omnomnomnom
Dec 10 '18 at 3:34
@72D do you understand the difference between $Bbb Z_2$ and $Bbb Z^2$?
– Omnomnomnom
Dec 10 '18 at 3:38
@Omnomnomnom, I edited but still I have same questions.
– 72D
Dec 10 '18 at 4:12
You seem to have misread the excerpt; the matrix corresponds to a homomorphism $Bbb Z^2 to Bbb Z^3$. Do you understand how this is different from what you typed? Does this clear anything up?
– Omnomnomnom
Dec 10 '18 at 3:30
You seem to have misread the excerpt; the matrix corresponds to a homomorphism $Bbb Z^2 to Bbb Z^3$. Do you understand how this is different from what you typed? Does this clear anything up?
– Omnomnomnom
Dec 10 '18 at 3:30
I'm confused with your use of subscripts vs what I think ought to be superscripts. Usually, $mathbb{Z_2}$ would be the integers mod $2$, where as $mathbb{Z^2}$ would be the direct product of the integers with itself, ie, pairs of integers where operations are done by component. Where you are using matrices, I'm assuming you actually want the latter.
– Josh B.
Dec 10 '18 at 3:31
I'm confused with your use of subscripts vs what I think ought to be superscripts. Usually, $mathbb{Z_2}$ would be the integers mod $2$, where as $mathbb{Z^2}$ would be the direct product of the integers with itself, ie, pairs of integers where operations are done by component. Where you are using matrices, I'm assuming you actually want the latter.
– Josh B.
Dec 10 '18 at 3:31
Also, it is not at all clear what you mean by $$ M=begin{pmatrix}1&3\2&3\5&9end{pmatrix} begin{pmatrix}1\0end{pmatrix} ne begin{pmatrix}1\2\0end{pmatrix} $$
– Omnomnomnom
Dec 10 '18 at 3:34
Also, it is not at all clear what you mean by $$ M=begin{pmatrix}1&3\2&3\5&9end{pmatrix} begin{pmatrix}1\0end{pmatrix} ne begin{pmatrix}1\2\0end{pmatrix} $$
– Omnomnomnom
Dec 10 '18 at 3:34
@72D do you understand the difference between $Bbb Z_2$ and $Bbb Z^2$?
– Omnomnomnom
Dec 10 '18 at 3:38
@72D do you understand the difference between $Bbb Z_2$ and $Bbb Z^2$?
– Omnomnomnom
Dec 10 '18 at 3:38
@Omnomnomnom, I edited but still I have same questions.
– 72D
Dec 10 '18 at 4:12
@Omnomnomnom, I edited but still I have same questions.
– 72D
Dec 10 '18 at 4:12
add a comment |
1 Answer
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In linear algebra, our $3 times 2$ matrix corresponds to the map
$$
pmatrix{a\b} mapsto pmatrix{1&3\2&3\5&9} pmatrix{a\b} = pmatrix{a + 3b\2a + 3b\5a+9b}
$$
This map is linear when $a,b$ are taken to be field elements, which is to say that a matrix of this shape traditionally represents a module homomorphism from $Bbb F^2$ to $Bbb F^3$. Similarly, when we take $a,b$ to be elements of $Bbb Z$, we end up with a module homomorphism from $Bbb Z^2$ to $Bbb Z^3$.
I'm not sure about the answer to your second question, but I suspect that somebody else might. The full excerpt is as follows:
this matrix corresponds to a homomorphism $Bbb Z^2 → Bbb Z^3$, hence to a $Bbb Z$-module; that is, a finitely generated abelian group $G$. The reader should figure out what $G$ is more explicitly (in terms of the classification of §IV.6, cf. Exercise 2.19) before reading on. (Aluffi, p. 345)
The referenced exercise:
Re-prove Corollary IV.6.5 as a corollary of Proposition 2.11 [viz. classification of finite abelian groups]. In fact, prove the more general fact that every finitely generated abelian group is a direct sum of cyclic groups. (Aluffi, p. 327)
answered Dec 10 '18 at 4:35
Omnomnomnom
126k788176
Perhaps the image of this homomorphism (which is a submodule of $Bbb Z^3$) is what Aluffi means by the $Bbb Z$-module corresponding to the homomorphism, but this is just my best guess.
– Omnomnomnom
Dec 10 '18 at 4:39
add a comment |
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In linear algebra, our $3 times 2$ matrix corresponds to the map
$$
pmatrix{a\b} mapsto pmatrix{1&3\2&3\5&9} pmatrix{a\b} = pmatrix{a + 3b\2a + 3b\5a+9b}
$$
This map is linear when $a,b$ are taken to be field elements, which is to say that a matrix of this shape traditionally represents a module homomorphism from $Bbb F^2$ to $Bbb F^3$. Similarly, when we take $a,b$ to be elements of $Bbb Z$, we end up with a module homomorphism from $Bbb Z^2$ to $Bbb Z^3$.
I'm not sure about the answer to your second question, but I suspect that somebody else might. The full excerpt is as follows:
this matrix corresponds to a homomorphism $Bbb Z^2 → Bbb Z^3$, hence to a $Bbb Z$-module; that is, a finitely generated abelian group $G$. The reader should figure out what $G$ is more explicitly (in terms of the classification of §IV.6, cf. Exercise 2.19) before reading on. (Aluffi, p. 345)
The referenced exercise:
Re-prove Corollary IV.6.5 as a corollary of Proposition 2.11 [viz. classification of finite abelian groups]. In fact, prove the more general fact that every finitely generated abelian group is a direct sum of cyclic groups. (Aluffi, p. 327)
answered Dec 10 '18 at 4:35
Omnomnomnom
126k788176
Perhaps the image of this homomorphism (which is a submodule of $Bbb Z^3$) is what Aluffi means by the $Bbb Z$-module corresponding to the homomorphism, but this is just my best guess.
– Omnomnomnom
Dec 10 '18 at 4:39
add a comment |
In linear algebra, our $3 times 2$ matrix corresponds to the map
$$
pmatrix{a\b} mapsto pmatrix{1&3\2&3\5&9} pmatrix{a\b} = pmatrix{a + 3b\2a + 3b\5a+9b}
$$
This map is linear when $a,b$ are taken to be field elements, which is to say that a matrix of this shape traditionally represents a module homomorphism from $Bbb F^2$ to $Bbb F^3$. Similarly, when we take $a,b$ to be elements of $Bbb Z$, we end up with a module homomorphism from $Bbb Z^2$ to $Bbb Z^3$.
I'm not sure about the answer to your second question, but I suspect that somebody else might. The full excerpt is as follows:
this matrix corresponds to a homomorphism $Bbb Z^2 → Bbb Z^3$, hence to a $Bbb Z$-module; that is, a finitely generated abelian group $G$. The reader should figure out what $G$ is more explicitly (in terms of the classification of §IV.6, cf. Exercise 2.19) before reading on. (Aluffi, p. 345)
The referenced exercise:
Re-prove Corollary IV.6.5 as a corollary of Proposition 2.11 [viz. classification of finite abelian groups]. In fact, prove the more general fact that every finitely generated abelian group is a direct sum of cyclic groups. (Aluffi, p. 327)
answered Dec 10 '18 at 4:35
Omnomnomnom
126k788176
Perhaps the image of this homomorphism (which is a submodule of $Bbb Z^3$) is what Aluffi means by the $Bbb Z$-module corresponding to the homomorphism, but this is just my best guess.
– Omnomnomnom
Dec 10 '18 at 4:39
add a comment |
In linear algebra, our $3 times 2$ matrix corresponds to the map
$$
pmatrix{a\b} mapsto pmatrix{1&3\2&3\5&9} pmatrix{a\b} = pmatrix{a + 3b\2a + 3b\5a+9b}
$$
This map is linear when $a,b$ are taken to be field elements, which is to say that a matrix of this shape traditionally represents a module homomorphism from $Bbb F^2$ to $Bbb F^3$. Similarly, when we take $a,b$ to be elements of $Bbb Z$, we end up with a module homomorphism from $Bbb Z^2$ to $Bbb Z^3$.
I'm not sure about the answer to your second question, but I suspect that somebody else might. The full excerpt is as follows:
this matrix corresponds to a homomorphism $Bbb Z^2 → Bbb Z^3$, hence to a $Bbb Z$-module; that is, a finitely generated abelian group $G$. The reader should figure out what $G$ is more explicitly (in terms of the classification of §IV.6, cf. Exercise 2.19) before reading on. (Aluffi, p. 345)
The referenced exercise:
Re-prove Corollary IV.6.5 as a corollary of Proposition 2.11 [viz. classification of finite abelian groups]. In fact, prove the more general fact that every finitely generated abelian group is a direct sum of cyclic groups. (Aluffi, p. 327)
answered Dec 10 '18 at 4:35
Omnomnomnom
126k788176
In linear algebra, our $3 times 2$ matrix corresponds to the map
$$
pmatrix{a\b} mapsto pmatrix{1&3\2&3\5&9} pmatrix{a\b} = pmatrix{a + 3b\2a + 3b\5a+9b}
$$
This map is linear when $a,b$ are taken to be field elements, which is to say that a matrix of this shape traditionally represents a module homomorphism from $Bbb F^2$ to $Bbb F^3$. Similarly, when we take $a,b$ to be elements of $Bbb Z$, we end up with a module homomorphism from $Bbb Z^2$ to $Bbb Z^3$.
I'm not sure about the answer to your second question, but I suspect that somebody else might. The full excerpt is as follows:
this matrix corresponds to a homomorphism $Bbb Z^2 → Bbb Z^3$, hence to a $Bbb Z$-module; that is, a finitely generated abelian group $G$. The reader should figure out what $G$ is more explicitly (in terms of the classification of §IV.6, cf. Exercise 2.19) before reading on. (Aluffi, p. 345)
The referenced exercise:
Re-prove Corollary IV.6.5 as a corollary of Proposition 2.11 [viz. classification of finite abelian groups]. In fact, prove the more general fact that every finitely generated abelian group is a direct sum of cyclic groups. (Aluffi, p. 327)
answered Dec 10 '18 at 4:35
Omnomnomnom
126k788176
answered Dec 10 '18 at 4:35
Omnomnomnom
126k788176
answered Dec 10 '18 at 4:35
Omnomnomnom
126k788176
answered Dec 10 '18 at 4:35
Omnomnomnom
126k788176
126k788176
Perhaps the image of this homomorphism (which is a submodule of $Bbb Z^3$) is what Aluffi means by the $Bbb Z$-module corresponding to the homomorphism, but this is just my best guess.
– Omnomnomnom
Dec 10 '18 at 4:39
add a comment |
Perhaps the image of this homomorphism (which is a submodule of $Bbb Z^3$) is what Aluffi means by the $Bbb Z$-module corresponding to the homomorphism, but this is just my best guess.
– Omnomnomnom
Dec 10 '18 at 4:39
Perhaps the image of this homomorphism (which is a submodule of $Bbb Z^3$) is what Aluffi means by the $Bbb Z$-module corresponding to the homomorphism, but this is just my best guess.
– Omnomnomnom
Dec 10 '18 at 4:39
Perhaps the image of this homomorphism (which is a submodule of $Bbb Z^3$) is what Aluffi means by the $Bbb Z$-module corresponding to the homomorphism, but this is just my best guess.
– Omnomnomnom
Dec 10 '18 at 4:39
add a comment |
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You seem to have misread the excerpt; the matrix corresponds to a homomorphism $Bbb Z^2 to Bbb Z^3$. Do you understand how this is different from what you typed? Does this clear anything up?
– Omnomnomnom
Dec 10 '18 at 3:30
I'm confused with your use of subscripts vs what I think ought to be superscripts. Usually, $mathbb{Z_2}$ would be the integers mod $2$, where as $mathbb{Z^2}$ would be the direct product of the integers with itself, ie, pairs of integers where operations are done by component. Where you are using matrices, I'm assuming you actually want the latter.
– Josh B.
Dec 10 '18 at 3:31
Also, it is not at all clear what you mean by $$ M=begin{pmatrix}1&3\2&3\5&9end{pmatrix} begin{pmatrix}1\0end{pmatrix} ne begin{pmatrix}1\2\0end{pmatrix} $$
– Omnomnomnom
Dec 10 '18 at 3:34
@72D do you understand the difference between $Bbb Z_2$ and $Bbb Z^2$?
– Omnomnomnom
Dec 10 '18 at 3:38
@Omnomnomnom, I edited but still I have same questions.
– 72D
Dec 10 '18 at 4:12